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Homework Help: The Sum of a series with exponents

  1. Dec 15, 2008 #1
    1. The problem statement, all variables and given/known data

    Problem: Indicate whether the series converges or diverges. If it converges, find its sum.

    THE SERIES:

    sum.JPG



    2. Relevant equations

    The ratio test and w/e equation is used to find the sum of this particular series

    3. The attempt at a solution

    I was able to find that the series converges, using the ratio test. However, I cannot find the sum of the series. I do not see any way in which I could manipulate the geometric series, or anything like that. Could someone please enlighten me on how to find the sum of this series, or any series in such a form.
     
  2. jcsd
  3. Dec 15, 2008 #2

    Dick

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    I don't think you need to do a lot of manipulation on a geometric series to find the sum. It should be pretty straightforward. Just factor out the k=1 term.
     
  4. Dec 15, 2008 #3
    I have factored out the k=1 term...yet, I still do not know exactly what to do. I wrote out the series for both the numerator and the denominator, yet I do not see what to do next.

    Here is my work:

    sumone.jpg

    Another hint at the problem solving process will be greatly appreciated.

    Thanks.
     
  5. Dec 15, 2008 #4

    rock.freak667

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    you could just write 2k+1=2k*2 and 5k-1=5k*5-1
     
  6. Dec 15, 2008 #5
    Thanks for all the hints....but now, I have another problem. I obtained an answer of 50/3...however, the solution booklet says the answer is 20/3...

    Here is my work:

    sumagain.jpg

    Is this a typo in the manual, or is my answer truly wrong? If so, please tell me where I messed up.
     
  7. Dec 16, 2008 #6

    gabbagabbahey

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    Gold Member

    The problem is that your sum is going from 1 to infinity, not zero to infinity. The formula for the geometric series that you used requires that the sum goes from zero to infinity:

    [tex]\sum_{k=0}^{\infty}ax^k=\frac{a}{1-x}[/tex]

    But

    [tex]\sum_{k=1}^{\infty}ax^k=\frac{a}{1-x}-a[/tex]
     
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